10.26 NOTE

P4742 [Wind Festival] Running In The Sky

题目传送门

思路

没啥营养，和所驼门王那一题一样，Tarjan 缩点，而后 DAG 上 DP。甚至还更简单一点。唯一需要注意的是要仔细考虑一下状态转移方程，这点很重要，不然会出大问题。

Code

#include<bits/stdc++.h>
#define Iseri namespace
#define Nina std
#define Kawaragi int
#define Momoka main
#define ls(p) (p<<1)
#define rs(p) (p<<1|1)
#define ll long long
#define ull unsigned long long
#define endl "\n"
#define pii pair<ll,ll>
const int maxn=200005;
const int inf=0x3f3f3f3f;

using Iseri Nina;

inline ll read(){
	ll x=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-48;ch=getchar();}
	return x*f;
}

//===========================================================

ll n,m,def[maxn],dfn[maxn],low[maxn],f[maxn],g[maxn],w[maxn],x,y,inst[maxn],t,cnt,rd[maxn];
ll mx[maxn],sg[maxn],ans=1;
vector<ll>v[maxn],e[maxn];
stack<ll>s;

inline void tarjan(ll u){
	dfn[u]=low[u]=++t;
	s.push(u);inst[u]=1;

	for(auto i:e[u]){
		if(!dfn[i]){
			tarjan(i);
			low[u]=min(low[u],low[i]);
		}else{
			if(inst[i])low[u]=min(low[u],dfn[i]);
		}
	}

	if(dfn[u]==low[u]){
		cnt++;
		while(s.top()!=u){
			ll tmp=s.top();s.pop();
			def[tmp]=cnt;
			sg[cnt]+=w[tmp];
			mx[cnt]=max(mx[cnt],w[tmp]);
			inst[tmp]=0;
		}
		s.pop();inst[u]=0;
		def[u]=cnt;sg[cnt]+=w[u];mx[cnt]=max(mx[cnt],w[u]);
	}
	return;
}

Kawaragi Momoka(){
	n=read(),m=read();
	for(ll i=1;i<=n;i++)w[i]=read();
	for(ll i=1;i<=m;i++){
		x=read(),y=read();
		e[x].push_back(y);
	}

	for(ll i=1;i<=n;i++)if(!dfn[i])tarjan(i);
	for(ll i=1;i<=n;i++){
		for(auto j:e[i]){
			if(def[i]!=def[j])v[def[i]].push_back(def[j]),rd[def[j]]++;
		}
	}

	queue<ll>q;
	for(ll i=1;i<=cnt;i++){
		f[i]=sg[i],g[i]=mx[i];
		if(rd[i]==0)q.push(i);
	}
	while(!q.empty()){
		ll u=q.front();q.pop();
		for(auto i:v[u]){
			if(f[u]+sg[i]>f[i]){
				f[i]=f[u]+sg[i];
				g[i]=max(g[u],mx[i]);
			}else if(f[u]+sg[i]==f[i])g[i]=max(g[i],g[u]);
			if(--rd[i]==0)q.push(i);
		}
	}

	for(ll i=2;i<=cnt;i++){
		if(f[i]>=f[ans]||(f[i]==f[ans]&&g[i]>g[ans]))ans=i;
	}
	printf("%lld %lld",f[ans],g[ans]);
	return 0;
}